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Luda [366]
4 years ago
5

1/2 ln(x + 3) - lnx = 0 solve for x

Mathematics
1 answer:
rusak2 [61]4 years ago
6 0
1

 

Simplify \frac{1}{2}\imath n(x+3)​2​​1​​ın(x+3) to \frac{\imath n(x+3)}{2}​2​​ın(x+3)​​

\frac{\imath n(x+3)}{2}-\imath nx=0​2​​ın(x+3)​​−ınx=0


2

 

Add \imath nxınx to both sides

\frac{\imath n(x+3)}{2}=\imath nx​2​​ın(x+3)​​=ınx


3

 

Multiply both sides by 22

\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2


4

 

Regroup terms

\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı


5

 

Cancel \imathı on both sides

n(x+3)=nx\times 2n(x+3)=nx×2


6

 

Divide both sides by nn

x+3=\frac{nx\times 2}{n}x+3=​n​​nx×2​​


7

 

Subtract 33 from both sides

x=\frac{nx\times 2}{n}-3x=​n​​nx×2​​−3


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Answer:

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Step-by-step explanation:

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I need help with this. Thanks <br><br><br> I need to find the degree
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Step-by-step explanation:

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_____

<em>Comment on using a calculator</em>

If you use the ATAN2( ) function of a graphing calculator or spreadsheet, it will give you the angle in the proper quadrant. If you use the arctangent function (tan⁻¹) of a typical scientific calculator, it will give you a 4th-quadrant angle when the ratio is negative. You must recognize that the desired 2nd-quadrant angle is 180° more than that.

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It may help you to consider looking at the "reference angle." In this geometry, it is the angle between the vector v and the -x axis. The coordinates tell you the lengths of the sides of the triangle vector v forms with the -x axis and a vertical line from that axis to the tip of the vector. Then the trig ratio you're interested in is ...

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This is the tangent of the reference angle, which will be ...

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You can see from your diagram that the angle CCW from the +x axis will be the supplement of this value, 180° -60.95° = 119.05°.

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